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4 months ago
magisterchessmutt
Hi once again Jim A., I've just had time to review your most extensive research efforts to fathom that Saxon/Indus Ft. here and applaud all your efforts in that regard. They match most of my own findings over the years also except now what most interests me in all this is just where and when this unit originated. Some would say of course Indus Valley however I am not so sure of that, but like
Forum: Mysteries
4 months ago
magisterchessmutt
Hi Jim A., Melissa et al, This is always an interesting speculation as far as intended measurement precision goes relevant to the Drusian or Saxon Ft. of presumed 1.1 ratio to the English 1 and equivalent digits. If the intended value of this Ft. was more like .3333 Meter of the ancient definition, it would be equivalent to 13.125 inches and not 13.2 inches. Again, how precise has this incredi
Forum: Mysteries
6 months ago
magisterchessmutt
Hi Jim A., I've been making a few more observations concerning that intriguing discrepancy between a half hekat volume sphere with circumference equal to an ERC, and it's cubic edge counterpart totaling 60 hekats. The reason a sq.rt. of 10 version of Pi works out equally, is due to the inverse of sq.rt. 10 being the same value as it's obverse with only the decimal place shifted by 10 of course
Forum: Mysteries
6 months ago
magisterchessmutt
<Nonetheless, 1 (royal cubit circumference)/square root 10 = .316227766..., divided by 2 = .15811388..., cubed = .00395284..., x 4/3 x square root 10 = .016666666..., x 30 = .5, or exactly 1/2 hekat for the sphere, with 30 hekat for the cubic cubit, with the reciprocal pi value for the diameter of the sphere of 1/square root 10.> Aha, good observation Jim A., If there's a proper Pi pa
Forum: Mysteries
6 months ago
magisterchessmutt
<That slight differential of .50660 and .5 appears to be fixed no matter what initial cubic ERC you employ. If I start off with an RC cubit edge of 8777.1428 cu.in at 20.6277 in. / Pi = 6.566 in. / 2 = 3.283 in. cubed at 35.38454 cu.in. x 4/3 x Pi = 148.2184 cu.in. for the half hekat x 60 = 8893.1 cu.in., and then divided by the initial input of 8777.1428 cu.in. = 1.01321 x .5 = .50660 just as
Forum: Mysteries
6 months ago
magisterchessmutt
<Given 30 hekat contained in the cubic royal cubit, as stated in the ancient Egyptian texts, this is probably a better way of calculating the accuracy of the 1/2 hekat sphere proposed in the article above: With a circumference of 1 royal cubit for the sphere, the diameter is 1/pi = .3183..., divided by 2 = .15915... for the radius. .15915... cubed = .00403..., x 4/3 x pi = .0168868... cubic
Forum: Mysteries
6 months ago
magisterchessmutt
Hi Melissa, I keep confusing myself trying to comprehend all the variables involved in the SOL/Lunar Geodetic formula. Essentially it's a dynamic process that specifies the Earth's dimensions are shaped by the gravitational forces acting upon it by the Sun and Moon which gives it an oblate spheroid shape instead of circular. I've already mentioned that the Moon's eccentric orbital parameters f
Forum: Mysteries
6 months ago
magisterchessmutt
<Perhaps this 573,750,000 figure could be (oh... I'm so predictable :o) lunar? 573,750,000 being 29.53125 x 408,000,000 / 21 So with 573,750,000 x 144/7 = 11,802,857,142.857 inches per second, you could think of the speed of light as: 29.53125 x 400,000,000 x 1224/1225 inches per second, since 816 x 144 x 500,000 / (21 x 7) = 400,000,000.> Hi Melissa, As usual you present more intr
Forum: Mysteries
6 months ago
magisterchessmutt
<Actually, to go back to the Hebrew calendar: Quote Irv Bromberg The Speed of Light The speed of light in a vacuum is exactly 299,792,458 metres per second of atomic time — this defines the standard length of the metre. A close approximation, accurate to within less than 0.07%, is 3/10 of a billion metres per second. There are 10/3 seconds per chelek so the speed of light per chelek ≈ 10/3
Forum: Mysteries
6 months ago
magisterchessmutt
<As I mentioned before, the trimmed English foot is considerably more geodetic, giving 25,000 miles for the polar circumference, and since the trimmed English foot times 11/10 equals the whole number of 18 digits, and because the Egyptian digit is equal to 1/100,000 of a minute of latitude, the Sumerian, Indus Valley and Northern feet are also more geodetic, to the extent that they are equal t
Forum: Mysteries
6 months ago
magisterchessmutt
<This is an interesting if somewhat confusing article about the various stages and rebuilds and restorations of the tower of Babel and additional cuneiform inscriptions about the tower and the Babylonian measurement system that have been discovered since the Smith tablet. George gives the length of the Old Babylonian cubit as 50 cm of 30 shusi and he also gives the length of the Neo-Babylonian
Forum: Mysteries
6 months ago
magisterchessmutt
<Is it possible that the reason 175/176 is so frequently applied to rationalize other ancient measurement systems in relation to the English inch, in terms of geodetic accuracy and actual measures of the lengths of the other ancient measures, is because there was an ancient English inch equal to 175/176 of the present English inch?> Hi Jim A., The more likely reason for the 176/175 ra
Forum: Mysteries
7 months ago
magisterchessmutt
<So certain forms of pi as lunar orbit ratios? And 864 has solar connections (sun diameter in miles, seconds in a solar day, etc), which would make 864/275 a solar pi... :) In any case, these pi variations become something more then than just workable easy ratios to use, for simple-minded people who couldn't work out actual pi, but intrinsically connected to the sun and moon, designed to fit
Forum: Mysteries
7 months ago
magisterchessmutt
Hi GHMB, Something caught my eye on a subject I have been studying for some time now regarding the ratio of gold to silver in early Babylonian and later Persian era up until the conquest of Alexander of Macedonia. The ratio of gold to silver was 1:13.333 or 40/3 for silver in ratio to gold. The question is why was this system established and upheld for so long when the recovery process of the
Forum: Mysteries
7 months ago
magisterchessmutt
Hi Melissa, Thanks for this very thoughtful and insightful post on the complexities involved with calculating longitude as opposed to latitude. I'm not as well read on the subject as I should be I think, but I have been aware that the Moon's circuit and phases played a large part in these calculations. I'm pretty sure the volumetric calculations involving the 8750 cu.in. Egyptian cubit and the
Forum: Mysteries
7 months ago
magisterchessmutt
<Also, considering the relation of 48/35, as in 48 digits = 35 imperial inches, the distance of 120 Roman feet, x 16 = 1920 digits, x 35/48 = 1400 imperial inches for the short side of the jugerum, giving 2800 inches for the long side. 1400/12 = 116.666... imperial feet for the short side, giving 233.333... feet for the long side, and 116.666... x 233.333... = 27222.222.... square feet. The sq
Forum: Mysteries
7 months ago
magisterchessmutt
Hi Jim A., Yes, an accurate timekeeping system would have to be employed in order to determine longitude degrees or some equivalent system. Knowing when a star rose at a particular point on the equator, and then determining the difference of it's rising at an earlier point east or later point west at a specific distance interval could resolve the matter perhaps. You are right in that it would
Forum: Mysteries
7 months ago
magisterchessmutt
<I agree with this. On pp 18-20 of my remen article, I propose the small pyramids from the 3rd dynasty that are scattered along the Nile from the delta to Elephantine as station points for a meridian survey, and the meridian distance from Elephantine to the northern limit of Egypt is 7.5 degrees.> Hi Jim A., There is plenty of evidence of surveys conducted throughout Egypt, but what a
Forum: Mysteries
7 months ago
magisterchessmutt
<39.375 English inches x 40,000,000 = 24,857.9545454 miles. Also, 35 English inches divided by 48 = .7291666 English inches (digit), and .7291666 x 20 x 99/70 = 20.625 English inches (royal cubit). So, whose survey was more accurate, c. 1700-1800 A.D. or c. 3000 B.C.?> Hi Jim A., There's really no excuse for this obvious correlation to be overlooked for so long now. 35 Englis
Forum: Mysteries
7 months ago
magisterchessmutt
<254400/360 = 706.666 stades per degree. I believe this is a result of the royal cubit being based on the remen times the square root of 2, giving an original measure of 300,000 remen per meridian degree, or 1000 remen stades of 300 remen per stade, or 360,000 remen stades in the circumference, which are all exactly accurate. A remen itr of 15,000 remen is 1/20 of a degree, and 7.5 degrees x 2
Forum: Mysteries
8 months ago
magisterchessmutt
Hi M and Jim A., That's a nice graphic which I'm sure will help others visualize this tricyclic relationship of ratios related to variant geodetic dimensions. The initial idea of a sphere of 25,000 Mi. circ. began long ago after reading John Michell's early works that suggested that figure. The idea developed to relate how the variant values of Pi as 25/8 = 3.125 could be involved in this conc
Forum: Mysteries
8 months ago
magisterchessmutt
Hi Jim A. and M, That is a good article on the Meter you referenced. I was sort of under the impression that piracy in the colonies had come to something of an abrupt ending at least by the 1720's, prompted by the execution of that most notorious pirate Edward Thatch aka Blackbeard in 1718, or at least in these parts of the colonies where I grew up and a lot of their ancestors still exist. Now
Forum: Mysteries
8 months ago
magisterchessmutt
<To go back to the 24,902.34375 mile equatorial circumference, it's still interesting to think of it as solar, since 24,902.34375 x 5,280 / (360 x 1,0000) = 365.234375. It's not a precise fit for the solar year, but intriguing. And as such the imperial system generally may represent a solar aspect of one great ancient measuring system. It would make sense then for the units derived from the me
Forum: Mysteries
8 months ago
magisterchessmutt
Hi Jim A. and all, There doesn't appear to be a direct path from the Canonical Circ. of John Michell's iconic production years ago, to the base figure of 25,000 Mi., but there are quite a few indirect paths as you have already pointed out, to which I would like to add one more. 25,000 Mi. x 224/225 x 4374/4375 = 24,883.2 Mi. The ever present ragisma comma that connects the two Polar Meridi
Forum: Mysteries
8 months ago
magisterchessmutt
Hi M, I'm not sure you are familiar with the Equatorial Circ. dimensions of the model I have been describing or the reason for the base value of 25,000 Mi. The reduction ratio to this figure of 255/256 = 24902.34375 Mi. x 5280 Ft. = 131484375 Ft. x 12 in. = 1,577,812,500 in. / 20.625 = 76,500,000 Egyptian cubits. This is in ratio to the Polar Meridian Circ. by 561/560 to obtain 1,575,000,000 i
Forum: Mysteries
8 months ago
magisterchessmutt
Hi M, You are right in saying there isn't a rational ratio discrepancy between a base unit of 25,000 Mi. and Michell's Canonical Circ. of 24,883.2 Mi. This model does have many uses and applications I've noted over years of studies of volumetric, weight, along with linear units. But it really doesn't seem to be a very useful meridian circ. value even if it can be used as such. Best regards,
Forum: Mysteries
8 months ago
magisterchessmutt
Hi Jim A., Well spotted on that 1750/1749 issue related to the discrepancy between the remen and cubit itr units. The only last thing that still bugs me is the rational ratio derivatives of a starting base of 25,000 Mi. in these calculations. All of the ones I have mentioned so far of reductions of 63/64 and 159/160 and also 175/176 and 255/256 to this base value in miles seems inherent in the
Forum: Mysteries
8 months ago
magisterchessmutt
<As I mentioned in a post above, and tried to explain in my remen article, I think the 99/100 discrepancy is due to Eratosthenes mistakenly thinking there were 700 stades of 300 royal cubits in a degree of latitude, when actually there were 707. I think the 707 is an artifact of the remen measure of 300,000 remen per degree, or 1000 stades of 300 remen, and 1000 / square root 2 for the royal c
Forum: Mysteries
8 months ago
magisterchessmutt
<Stephen, thanks for your response. Given a digit of .729 inches, a foot of 16 digits and a cubit of 24 digits: cubit/foot = 3/2 meter/cubit = 9/4 = (3/2) squared meter/foot = 27/8 = (3/2) cubed <<What I wonder the most about is how this simple relationship ever became confused to begin with.>> Agreed.> Hi Jim A., Agreed in principal yes, but I'm still wonde
Forum: Mysteries
8 months ago
magisterchessmutt
<From the passage you cited of Letronne's award winning paper from 1816 that got him admitted to the Academy, but was never published during his lifetime, he recognized that 250,000 cubits based on 1 1/2 times the length of the Roman foot was an exact measure of one degree of latitude, and 250,000 x 360 degrees = 90,000,000 cubits for the full circumference. Because Letronne died before Petrie
Forum: Mysteries
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